Given the functions $f(x) = x - 8$ and $g(x) = x^2 + 6x$, we need to find the function $h$ such that $h(x) = f(g(x))$, and then express the inverse function $h^{-1}(x)$.

First, let's find $h(x)$: $h(x) = f(g(x))$ $g(x) = x^2 + 6x$ $f(x) = x - 8$

So, $h(x) = f(g(x)) = f(x^2 + 6x) = (x^2 + 6x) - 8$ $h(x) = x^2 + 6x - 8$

Next, we need to find the inverse function $h^{-1}(x)$. To do this, we solve the equation $y = h(x)$ for $x$ in terms of $y$.

$y = x^2 + 6x - 8$

Rearrange the equation: $x^2 + 6x - (y + 8) = 0$

This is a quadratic equation in the form $ax^2 + bx + c = 0$. We can solve for $x$ using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, $a = 1$, $b = 6$, and $c = -(y + 8)$.

$x = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot (-(y + 8))}}{2 \cdot 1}$ $x = \frac{-6 \pm \sqrt{36 + 4(y + 8)}}{2}$ $x = \frac{-6 \pm \sqrt{36 + 4y + 32}}{2}$ $x = \frac{-6 \pm \sqrt{68 + 4y}}{2}$ $x = \frac{-6 \pm 2\sqrt{17 + y}}{2}$ $x = -3 \pm \sqrt{17 + y}$

Since the domain of $h$ is $\{x : x \geq -3\}$, we need to select the positive branch of the solution: $x = -3 + \sqrt{17 + y}$

Thus, the inverse function $h^{-1}(x)$ is: $h^{-1}(x) = -3 + \sqrt{17 + x}$

Would you like more details or have any questions? Here are some related questions you might be interested in:

1. How do you find the domain of a composite function?
2. What is the quadratic formula and how is it derived?
3. How do you verify if a function is invertible?
4. Can you provide an example of another composite function and its inverse?
5. What are the properties of inverse functions?

Tip: When solving for the inverse of a function, always check the domain and range to ensure the inverse is properly defined.